Question

if b^2=ac and 2b/1-b^2=a+c/1-ac then prove that a=b=c​

Answer 1

Answer:

Given,

=> a + c = 2b

=> a = 2b - c ---------- ( 1 )

And,

=> ( 1/b ) + ( 1/d) = 2/c

Taking L.C.M of b and d.

=> ( d + b )/bd = 2/c

=> ( b + d ) = 2bd/c

=> b = ( 2bd / c ) - d

=> b = ( 2bd - dc )/c

Taking out d as common,

=> b = d ( 2b - c )/ c -------- ( 2 )

Now, by divding ( 1 ) by ( 2 ),

=> a/b = ( 2b - c ) ÷ { d( 2b - c ) /c }

=> a/b = { ( 2b - c ) × c } ÷ { d ( 2b - c ) }

=> a/b = c/d

Answer 2

Answer:

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